Timeline for How to interpolate a height-map with normals
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Feb 9, 2012 at 4:14 | comment | added | Nathan Reed | I added code for interpolating along X. Changing it to interpolate along Z is left as an exercise for the reader. ;) | |
| Feb 9, 2012 at 4:13 | history | edited | Nathan Reed | CC BY-SA 3.0 |
added example code
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| Feb 7, 2012 at 0:23 | comment | added | Felix K. | Hmm but it's a heightmap with fixed points, so 3D interpolation does not makes any sense to me. ;-) Anyway i gonna test it tomorrow, if it's right and you updated your answer the credits belong to you. | |
| Feb 7, 2012 at 0:18 | comment | added | Nathan Reed | Right. You'd do the interpolation in 3D, at least that's what makes sense to me. You could project it down to 2D by throwing out one of the coordinates - for the X-tangent, throw out the Z value, and vice versa. If you need a slope instead of a tangent, divide: Y/X for the X-tangent, Y/Z for the Z-tangent. | |
| Feb 6, 2012 at 23:40 | comment | added | Felix K. | But then i still have a 3D coordinate, not a single float. :-/ | |
| Feb 6, 2012 at 23:02 | comment | added | Nathan Reed | Oh. You just rotate it 90 degrees. I guess you're using a Y-up coordinate system, so take the normal and rotate by -90 around Z to get the X-tangent, and by 90 around X to get the Z-tangent. | |
| Feb 6, 2012 at 22:26 | comment | added | Felix K. | Thanks for the suggestion, but my main problem is still how to get the tangent from the 3D-normal. | |
| Feb 6, 2012 at 22:16 | history | answered | Nathan Reed | CC BY-SA 3.0 |